If the identifying information on each license plate consists of (letter-1)(letter-2)(digit-1)(digit-2)(digit-3)(digit-4), and repetition is allowed: letter-1 has 26 choices, and for each one ... letter-2 has 26 choices, and for each one ... digit-1 has 10 choices, and for each one ... digit-2 has 10 choices, and for each one ... digit-3 has 10 choices, and for each one ... digit-4 has 10 choices. Total number of choices = 262 x 104 = ( 26 x 26 x 10 x 10 x 10 x 10 ) = 676 x 10,000 = 6,760,000
the repeated digits can lie in any of 3 positions:XXYZ,YXXZ, or YZXXTherefore, the number of different possible repeated digit combinations is 10×3=30 (there are 10 different possible digits).There are then 9 values left that Y could be, after the X's have their values, and 8 values left that Z could be, to ensure no unintended repeated numbers.This gives (10×3)×9×8=2160. There are therefore 2160 possible codes.
There are 4 possible numbers if the digits are not repeated; 18 if they are. Those are 3-digit numbers, assuming that zero would not be a leading digit. If zero is allowed for a leading digit, then you can have 6 for the non repeated, and 27 if repetition.
This permutation problem depends on whether the numbers are allowed to be repeated. If they are, there are a possible 9999 numbers, starting with 0001 and running sequentially to 9999. If they are not allowed to be repeated, there are a possible 5040 combinations.
It is possible to tessellate a plane with squares, triangles, and hexagons. To tessellate something means to cover it with repeated use of a single shape, without gaps or overlapping.
0.225 Repeated, 0.225, 0.25 Repeated and 0.25.
There are 26 possible letters and 10 possible numbers. The number of license plates possible is then 26*26*10*10*10*10 = 6760000.
If you include 0000, ten thousand unique four digit codes are possible.
10*9*8*7*6=30,240
If the numbers and letters can be repeated then there are 45,697,600 possible outcomes. If the letters and numbers can not be repeated there are 32,292,000 possible outcomes.
If the numbers can be repeated and the numbers are 0-9 then there are 1000 different combinations.
9x8x7x6x5x4x3x2x1 or 9! which equals 362880 possible combinations if no digits are repeated
the repeated digits can lie in any of 3 positions:XXYZ,YXXZ, or YZXXTherefore, the number of different possible repeated digit combinations is 10×3=30 (there are 10 different possible digits).There are then 9 values left that Y could be, after the X's have their values, and 8 values left that Z could be, to ensure no unintended repeated numbers.This gives (10×3)×9×8=2160. There are therefore 2160 possible codes.
a repeated word followed by random numbers.... there is no way a person can answer that with so little information
263 X 103 = 17576000
Number of possible groups of 3 letters = 26 x 25 x 24 = 15,600. For each of these . . .Number of possible groups of 3 digits = 9 x 9 x 8 = 648 .Total number of possible distinct plates = 15,600 x 648 = 10,108,800
Parallelism.
Any three letters, in any order, including repeated letters gives 263 combinations each of which could have one of 9 digits so 26 x 26 x 26 x 9 ie 158184 different plates.